arthham

JEE Advanced 2023 · Three Dimensional Geometry

Let ℓ1\ell_1 and ℓ2\ell_2 be the lines r⃗1=λ(i^+j^+k^)\vec{r}_1 = \lambda(\hat{i} + \hat{j} + \hat{k}) and r⃗2=(j^−k^)+μ(i^+k^)\vec{r}_2 = (\hat{j} - \hat{k}) + \mu(\hat{i} + \hat{k}), respectively. Let X be the set of all the planes H that contain the line ℓ1\ell_1. For a plane H, let d(H) denote the smallest possible distance between the points of ℓ2\ell_2 and H. Let H0H_0 be a plane in X for which d(H0H_0) is the maximum value of d(H) as H varies over all planes in X. Match each entry in List-I to the correct entries in List-II. | List-I | List-II | | --- | --- | | (P) The value of d(H0)d(H_0) is | (1) 3\sqrt{3} | | (Q) The distance of the point (0,1,2)(0, 1, 2) from H0H_0 is | (2) 13\frac{1}{\sqrt{3}} | | (R) The distance of origin from H0H_0 is | (3) 00 | | (S) The distance of origin from the point of intersection of planes y=zy = z, x=1x = 1 and H0H_0 is | (4) 2\sqrt{2} | | | (5) 12\frac{1}{\sqrt{2}} |

Answer: B. (P) →\to (5), (Q) →\to (4), (R) →\to (3), (S) →\to (1)

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Let ell₁ and ell₂ be the lines r⃗₁ = λ(î + ĵ + k̂) and r⃗₂ = (ĵ - k̂) + μ(î + k̂), respectively. Let X be the set of all the planes H that contain the line ell₁. For a plane H, let d(H) denote the smallest possible distance between the points of ell₂ and H. Let H₀ be a plane in X for which d(H₀) is the maximum value of d(H) as H varies over all planes in X. Match each entry in List-I to the correct entries in List-II. | List-I | List-I ℓ1:r⃗1=λ(i^+j^+k^)\ell_1: \vec{r}_1 = \lambda(\hat{i} + \hat{j} + \hat{k}) ℓ2:r⃗2=(j^−k^)+μ(i^+k^)\ell_2: \vec{r}_2 = (\hat{j} - \hat{k}) + \mu(\hat{i} + \hat{k}) X={H:ℓ1⊂H},d(H)=min⁡pt∈ℓ2dist(pt,H)X = \{H : \ell_1 \subset H\}, \quad d(H) = \min_{\text{pt} \in \ell_2} \text{dist}(\text{pt}, H) H0∈X: d(H0)=max⁡H∈Xd(H)H_0 \in X: \ d(H_0) = \max_{H \in X} d(H) To do: match List-I with List-II\text{To do: match List-I with List-II} ℓ1:r⃗1=λ(i^+j^+k^),a⃗=(1,1,1)\ell_1: \vec{r}_1 = \lambda(\hat{i}+\hat{j}+\hat{k}), \quad \vec{a}=(1,1,1) ℓ2:r⃗2=(0,1,−1)+μ(1,0,1),b⃗=(1,0,1)\ell_2: \vec{r}_2 = (0,1,-1) + \mu(1,0,1), \quad \vec{b}=(1,0,1) H∈X:n⃗⋅r⃗=0,n⃗⋅(1,1,1)=0H \in X: \vec{n}\cdot\vec{r}=0, \quad \vec{n}\cdot(1,1,1)=0 ⇒n1+n2+n3=0\Rightarrow n_1 + n_2 + n_3 = 0 Plan: 1. d(H)=dist(ℓ2,H)\text{Plan: } 1.\ d(H)=\text{dist}(\ell_2,H) 2. max⁡d(H)⇒n⃗⇒H02.\ \max d(H) \Rightarrow \vec{n} \Rightarrow H_0 3. the four distances (P),(Q),(R),(S)3.\ \text{the four distances } (P),(Q),(R),(S) d(H)=0  if  b⃗⋅n⃗≠0d(H) = 0 \ \text{ if } \ \vec{b}\cdot\vec{n} \neq 0 d(H)=∣n⃗⋅B⃗∣∣n⃗∣  if  b⃗⋅n⃗=0d(H) = \dfrac{|\vec{n}\cdot\vec{B}|}{|\vec{n}|} \ \text{ if } \ \vec{b}\cdot\vec{n} = 0 b⃗⋅n⃗=0⇒n1+n3=0\vec{b}\cdot\vec{n}=0 \Rightarrow n_1 + n_3 = 0 n1+n2+n3=0⇒n2=0, n3=−n1n_1+n_2+n_3=0 \Rightarrow n_2=0,\ n_3=-n_1 n⃗∝(1,0,−1)\vec{n} \propto (1,0,-1) H0:x−z=0⇒x=zH_0: x - z = 0 \Rightarrow x = z d(H0)=∣(1,0,−1)⋅(0,1,−1)∣2=12d(H_0) = \dfrac{|(1,0,-1)\cdot(0,1,-1)|}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} (P)→(5)(P) \to (5) dist((0,1,2), x−z=0)=∣0−2∣2=2\text{dist}\big((0,1,2),\ x-z=0\big) = \dfrac{|0-2|}{\sqrt{2}} = \sqrt{2} (Q)→(4)(Q) \to (4) dist(O,H0)=0(O∈H0)\text{dist}(O, H_0) = 0 \quad (O \in H_0) (R)→(3)(R) \to (3) y=z, x=1, x=z⇒(1,1,1)y=z,\ x=1,\ x=z \Rightarrow (1,1,1) dist(O,(1,1,1))=3\text{dist}\big(O,(1,1,1)\big) = \sqrt{3} (S)→(1)(S) \to (1) (P)→(5), (Q)→(4), (R)→(3), (S)→(1)(P)\to(5),\ (Q)\to(4),\ (R)\to(3),\ (S)\to(1) ⇒option (2)\Rightarrow \text{option } (2)

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